Mathematics
In the given figure, AB = AC; D is the mid-point of BC; DP ⊥ BA and DQ ⊥ CA. Prove that:
(i) DP = DQ
(ii) AP = AQ
(iii) AD bisects ∠A
Answer
(i) In △ABC,
⇒ AB = AC (Given)
∴ ∠B = ∠C (Angles opposite to equal sides in a triangle are equal)
In △PDB and △QDC,
⇒ BD = CD (D is mid-point of BC)
⇒ ∠B = ∠C (Proved above)
⇒ ∠P = ∠Q (Both equal to 90°)
∴ △PDB ≅ △QDC (By A.A.S axiom)
⇒ DP = DQ (Corresponding parts of congruent triangles are equal)
Hence, proved that DP = DQ.
(ii) Since, △PDB ≅ △QDC
∴ BP = QC = y (let) (Corresponding parts of congruent triangles are equal)
⇒ AB = AC = x (let)
From figure,
⇒ AP = AB - BP = x - y …(1)
⇒ AQ = AC - QC = x - y …(2)
From eq.(1) and (2) we have :
∴ AP = AQ.
Hence, proved that AP = AQ.
(iii) In △ABD and △ACD,
⇒ AB = AC (Given)
⇒ BD = CD (Given)
⇒ AD = AD (Common side)
∴ △ABD ≅ △ACD (By S.S.S axiom)
⇒ ∠BAD = ∠CAD (Corresponding parts of congruent triangles are equal)
Hence, proved that AD bisects ∠A.
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