It is given that △ABC ≅ △RPQ. Is it true to say that BC = QR? Why?
Answer
Given, △ABC ≅ △RPQ.
It means that A ↔ R, B ↔ P, C ↔ Q, therefore, BC = PQ (corresponding sides are equal).
Hence, BC ≠ QR as they are not corresponding sides.
"If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent." Is the statement true? Why?
Answer
Given, statement "If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent."
The statement is not true as the angles must be included angles.
In the adjoining figure, AB = AC and AP = AQ. Prove that
(i) △APC ≅ △AQB
(ii) CP = BQ
(iii) ∠APC = ∠AQB.
Answer
(i) In △APC and △AQB we have,
AB = AC (Given)
AP = AQ (Given)
∠PAC = ∠QAB (Common angles)
Hence, by SAS axiom △APC ≅ △AQB.
(ii) As, △APC ≅ △AQB.
We know that corresponding sides of congruent triangles are equal.
∴ CP = BQ.
Hence, proved that CP = BQ.
(iii) As, △APC ≅ △AQB.
We know that corresponding angles of congruent triangles are equal.
∴ ∠APC = ∠AQB.
Hence, proved that ∠APC = ∠AQB.
In the adjoining figure, AB = AC, P and Q are points on BA and CA respectively such that AP = AQ. Prove that
(i) △APC ≅ △AQB
(ii) CP = BQ
(iii) ∠ACP = ∠ABQ.
Answer
(i) In △APC and △AQB we have,
AP = AQ (Given)
AB = AC (Given)
∠PAC = ∠QAB (Vertically opposite angles)
Hence, by SAS axiom △APC ≅ △AQB.
(ii) As, △APC ≅ △AQB.
We know that corresponding sides of congruent triangles are equal.
∴ CP = BQ.
Hence, proved that CP = BQ.
(iii) As, △APC ≅ △AQB.
We know that corresponding angles of congruent triangles are equal.
∴ ∠ACP = ∠ABQ.
Hence, proved that ∠ACP = ∠ABQ.
In the adjoining figure, AD = BC and BD = AC. Prove that:
∠ADB = ∠BCA and ∠DAB = ∠CBA.
Answer
In △ABD and △BAC we have,
AD = BC (Given)
BD = AC (Given)
AB = AB (Common sides)
∴ △ABD ≅ △BAC. (By SSS axiom)
We know that corresponding angles of congruent triangles are equal.
∴ ∠ADB = ∠BCA and ∠DAB = ∠CBA.
In the adjoining figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(i) △ABD ≅ △BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC
Answer
(i) In △ABD and △BAC,
AD = BC (Given)
∠BAD = ∠ABC (Given)
AB = AB (Common sides)
Hence, by SAS axiom △ABD ≅ △BAC.
(ii) As, △ABD ≅ △BAC.
We know that corresponding sides of congruent triangles are equal.
∴ BD = AC.
Hence, proved that BD = AC.
(iii) As, △ABD ≅ △BAC.
We know that corresponding angles of congruent triangles are equal.
∴ ∠ABD = ∠BAC.
Hence, proved that ∠ABD = ∠BAC.
In the adjoining figure, AB = DC and AB || DC. Prove that AD = BC.
Answer
In △ABD and △CDB,
AD = BC (Given)
∠ABD = ∠CDB (Alternate angles are equal)
BD = BD (Common sides)
∴ △ABD ≅ △CDB. (By SAS axiom)
We know that corresponding sides of congruent triangles are equal.
∴ AD = BC.
Hence, proved that AD = BC.
In the adjoining figure AC = AE, AB = AD and ∠BAD = ∠CAE. Show that BC = DE.
Answer
Join DE.
Given,
⇒ ∠BAD = ∠CAE
∴ ∠BAD + ∠DAC = ∠CAE + ∠DAC
⇒ ∠BAC = DAE.
In △ABC and △ADE,
AC = AE (Given)
AB = AD (Given)
∠BAC = DAE (Proved)
∴ △ABC ≅ △ADE by SAS axiom.
We know that corresponding sides of congruent triangles are equal.
∴ BC = DE.
Hence, proved that BC = DE.
In the adjoining figure, AB = AC and D is the midpoint of BC. Use SSS rule of congruency to show that
(i) △ABD ≅ △ACD
(ii) AD is bisector of ∠A
(iii) AD is perpendicular to BC.
Answer
(i) In △ABD and △ACD,
Given,
AB = AC (Given)
BD = CD (As D is the midpoint of BC)
AD = AD (Common)
Hence, by SSS axiom △ABD ≅ △ACD.
(ii) Since, △ABD ≅ △ACD
We know that corresponding angles of congruent triangle are equal.
∴ ∠BAD = ∠CAD.
Hence, proved that AD is bisector of ∠A.
(iii) △ABD ≅ △ACD
We know that corresponding angles of congruent triangles are equal.
∠ADB = ∠ADC.
Let ∠ADB = ∠ADC = x.
We know that,
⇒ ∠ADB + ∠ADC = 180°
⇒ x + x = 180°
⇒ 2x = 180°
⇒ x = 90°.
Hence, ∠ADB = ∠ADC = 90°.
Hence, proved that AD is perpendicular to BC.
In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that
(i) △ACE ≅ △DBF
(ii) AE = DF.
Answer
(i) In △ACE and △DBF,
Given,
AB = CD
⇒ AB + BC = CD + BC
⇒ AC = BD.
∠ACE = ∠DBF (Given)
CE = BF (Given)
Hence, by SAS axiom △ACE ≅ △DBF.
(ii) We know that, △ACE ≅ △DBF.
We know that corresponding sides of congruent triangles are equal.
∴ AE = DF.
Hence, proved that AE = DF.
Two line segments AC and BD bisect each other at P. Draw the diagram and prove that
(i) AB = CD
(ii) ∠BAC = ∠DCA
Answer
(i) Given, AC and BD bisect each other at P.
Join CD, BC, AB and AD.
In △BPA and △CPD,
As, P bisects AC and BD
⇒ PA = PC (P bisects AC and BD)
⇒ PB = PD (P bisects AC and BD)
⇒ ∠BPA = ∠CPD (Vertically opposite angles are equal).
∴ △BPA ≅ △CPD by SAS axiom.
We know that corresponding sides of congruent triangles are equal.
∴ AB = CD (By C.P.C.T.C.)
Hence, proved that AB = CD.
(ii) As proved in part (i),
△BPA ≅ △CPD by SAS axiom.
We know that corresponding angles of congruent triangles are equal.
∠DCP = ∠PAB ........................(1)
From figure we get,
∠DCP = ∠DCA and ∠PAB = ∠BAC.
Substituting above values in equation (1) we get,
∠DCA = ∠BAC.
Hence, proved that ∠BAC = ∠DCA.
Prove that the median drawn from the vertex P of an isosceles triangle △PQR with PQ = PR is perpendicular to QR and bisects ∠P.
Answer
Let PQR be an isosceles triangle with PQ = PR.
Draw a median from vertex P to the side QR. Let this median be PS, where S is the midpoint of QR.
In △PQS and △PRS,
⇒ PQ = PR (△PQR is an isosceles triangle).
⇒ QS = RS (Since PS is the median, S is the midpoint of QR).
⇒ PS = PS (Common side to both triangles).
By the SSS (Side-Side-Side) congruence criterion,
△PQS ≅ △PRS
We know that,
Corresponding parts of congruent triangles are congruent.
⇒ ∠PSQ = ∠PSR = x (let)
From figure,
⇒ ∠PSQ + ∠PSR = 180° [Linear pair]
⇒ x + x = 180°
⇒ 2x = 180°
⇒ x = = 90°.
⇒ ∠PSQ = ∠PSR = 90°.
Thus, PS is perpendicular to QR.
So, PS ⊥ QR.
⇒ ∠QPS = ∠RPS (By C.P.C.T.C.)
Thus, PS bisects ∠P.
Hence, the median drawn from the vertex P of an isosceles triangle △PQR is perpendicular to QR and bisects ∠P.
In the adjoining figure, find the values of x and y.
Answer
In △ABD and △CBD,
Given,
AB = BC (Given)
AD = CD (Given)
BD = BD (Common)
Hence, by SSS axiom △ABD ≅ △CBD.
We know that corresponding angles of congruent triangles are equal.
∴ (y + 5)° = 46° and (2x + 5)° = 35°
y° = 41° and 2x° = 30°
y = 41 and x = 15.
Hence, x = 15 and y = 41.