Mathematics
Answer
In △ADE,
AD = AE
⇒ ∠ADE = ∠AED (Angles opposite to equal sides in atriangle are equal)
From figure,
⇒ ∠ADE + ∠ADB = 180° (Linear pair)
⇒ ∠ADB = 180° - ∠ADE ….(1)
⇒ ∠AED + ∠AEC = 180° (Linear pair)
⇒ ∠AEC = 180° - ∠AED
⇒ ∠AEC = 180° - ∠ADE ….(2) (∵ ∠ADE = ∠AED)
From eq.(1) and (2), we have:
⇒ ∠AEC = ∠ADB
In △ABD and △ACE,
⇒ AD = AE (Given)
⇒ ∠ADB = ∠AEC (Proved above)
⇒ ∠BAD = ∠CAE (Given)
∴ △ABD ≅ △ACE (By A.S.A axiom)
⇒ AB = AC (Corresponding parts of congruent triangles are equal)
Hence, proved that AB = AC.
Related Questions
In an isosceles triangle, prove that the altitude from the vertex bisects the base.
If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles.
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
In the given figure, AB = AC. If BO and CO, the bisectors of ∠B and ∠C respectively meet at O and BC is produced to D, prove that ∠BOC = ∠ACD.
