Question

Q26. AD is an altitude of an isosceles triangle ABC in which AB=AC. Show th
I) triangle Abe congurant to triangle acf
ii) AD bisects angle a​

Answer 1

Correct Question :

AD is an altitude of an isosceles triangle ABC in which AB=AC. Show th

i) ∆ADB ≅ ∆ADC

ii) AD bisects ∠A

▣ Diagram :

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\linethickness{0.5mm}\qbezier(1, 0)(1,0)( 3,3)\qbezier(3, 0)(3,0)( 3,3)\qbezier(5,0)(5, 0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.5,0){\dashbox{1}(1,0.5)}\put(2.9,3.2){ \sf A }\put(0.6, - 0.2){ \sf B }\put(5.1, - 0.3){ \sf C }\put(2.9, - 0.5){ \sf D }\end{picture}$

▣ Given:

• AD is an altitude of an isosceles triangle.
• Therefore, AB = BC

▣ To Prove:

• i) ∆ADB ≅ ∆ADC

• ii) AD bisects ∠A

▣ Proof :

i) In ∆ADB and In ∆ADC

➝ AB = AC ......[Given]

➝ AD = AD ......[Common]

ஃ ∆ADB ≅ ∆ADC .......[RHS rule]

________________....

ii) $\because\:$ ∆ADB ≅ ∆ADC

⛬ ∠BAD = ∠CAD .......[CPCT]

⛬ AD bisects ∠A