Q19 of 47 Page 7

If two isosceles triangles have a common base, prove that the line joining their vertices bisects them at right angles.

Given: AB = AC and DB = DC


To prove: AD⊥ BC


Proof:



In Δ ABD and Δ ACD we have,


AB = AC (given)


BD = CD (given)


AD = AD (common)


By SSS congruency,


Δ ABD ≅ Δ ACD


⇒ ∠1 = ∠2 (By CPCT)


In Δ ABE and Δ ACE we have,


AB = AC (given)


∠1 = ∠2


AE = AE (common)


By SAS congruency,


Δ ABE ≅ Δ ACE


⇒ BE = CE (BY CPCT)


And ∠3 = ∠4


Also ∠3 + ∠4 = 180° (linear pair)


⇒ 2∠3 = 180°


⇒ ∠3 = 90°


⇒ ∠3 =∠4 = 90°


Hence AD bisects BC at 90°.


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