In the given figure, ABCD is a rectangle and X and Y are the mid-points of the sides DC and AB respectively. P and Q are the points of AD and BC respectively such that DP = BQ.
Prove that △ APX ≅ △ CQY.

Question

In the given figure, ABCD is a rectangle and X and Y are the mid-points of the sides DC and AB respectively. P and Q are the points of AD and BC respectively such that DP = BQ.
Prove that △ APX ≅ △ CQY.

KnowledgeBoat · Accepted Answer

Given, ABCD is a rectangle. X and Y are midpoints of DC and AB respectively. P and Q are points on AD and BC respectively such that DP = BQ Comparing sides AX and CY, In rectangle, opposite sides are equal, so AB = CD. Since X and Y are mid-points: DX = YB = Therefore, DX = YB........(1) Consider △ ADX and △ CBY, ⇒ AD = CB (opposite sides of a rectangle) ⇒ ∠D = ∠B = 90 ⇒ DX = YB (Proven above) So, △ ADX ≅ △ CBY (By S.A.S axiom) ∴ AX = CY (By C.P.C.T.C.) ........(2) Comparing sides AP and CQ, Given, AD = BC and DP = BQ. AD - DP = BC - BQ ∴ AP = CQ .......(3) Consider △ DPX and △ BQY: ⇒ DP = BQ (given) ⇒ ∠D = ∠B = 90 ⇒ DX = BY (Proved above) So, △ DPX ≅ △ BQY (By S.A.S axiom) ∴ PX = QY ..........(4) Consider △ APX and △ CQY. ⇒ AX = CY (From equation (2)) ⇒ AP = CQ (From equation (3)) ⇒ PX = QY (From equation (4)) Since all the three corresponding sides are equal, ∴ △ APX ≅ △ CQY (By S.S.S axiom) Hence proved that s△ APX ≅ △ CQY.

Mathematics

In the given figure, ABCD is a rectangle and X and Y are the mid-points of the sides DC and AB respectively. P and Q are the points of AD and BC respectively such that DP = BQ.
Prove that △ APX ≅ △ CQY.

Triangles

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Mathematics

In the given figure, ABCD is a rectangle and X and Y are the mid-points of the sides DC and AB respectively. P and Q are the points of AD and BC respectively such that DP = BQ.Prove that △ APX ≅ △ CQY.

Triangles

Related Questions

In △ ABC, AB = AC and the bisectors of angles B and C intersect at point O. Prove that :(i) BO = CO(ii) AO bisects angle BAC.

In the following figure, AB = EF, BC = DE and ∠B = ∠E = 90°. Prove that AD = FC.

A point O is taken inside a rhombus ABCD such that its distances from the vertices B and D are equal. Show that AOC is a straight line.

In the given figure, ABCD is a rectangle and X and Y are the mid-points of the sides DC and AB respectively. P and Q are the points of AD and BC respectively such that DP = BQ.
Prove that △ APX ≅ △ CQY.

In △ ABC, AB = AC and the bisectors of angles B and C intersect at point O. Prove that :
(i) BO = CO
(ii) AO bisects angle BAC.