Advertisements
Advertisements
प्रश्न
In the given figure; AB = BC and AD = EC.
Prove that: BD = BE.
उत्तर
In ΔABC,
AB = BC .......(given)
⇒ ∠BCA = ∠BAC .......(Angles opposite to equal sides are equal)
⇒ ∠BCD = ∠BAE ….(i)
Given, AD = EC
⇒ AD + DE = EC + DE ...(Adding DE on both sides)
⇒ AE = CD .....….(ii)
Now, in triangles ABE and CBD,
AB = BC .....(given)
∠BAE = ∠BCD ....[From (i)]
AE = CD ......[From (ii)]
⇒ ΔABE ≅ ΔCBD
⇒ BE = BD ....(cpct)
APPEARS IN
संबंधित प्रश्न
In the figure, given below, AB = AC.
Prove that: ∠BOC = ∠ACD.
Calculate x :
Calculate x :
Prove that a triangle ABC is isosceles, if: bisector of angle BAC is perpendicular to base BC.
In the given figure, AD = AB = AC, BD is parallel to CA and angle ACB = 65°. Find angle DAC.
In triangle ABC; AB = AC and ∠A : ∠B = 8 : 5; find angle A.
Using the information given of the following figure, find the values of a and b.
If the equal sides of an isosceles triangle are produced, prove that the exterior angles so formed are obtuse and equal.
ABC is a triangle. The bisector of the angle BCA meets AB in X. A point Y lies on CX such that AX = AY.
Prove that: ∠CAY = ∠ABC.
The bisectors of the equal angles B and C of an isosceles triangle ABC meet at O. Prove that AO bisects angle A.
