Question

ABC be a triangle on which D is a point on BC such that BD = CD. From D, the perpendiculars are drawn to the sides of the triangle. If the perpendicular are equal them to prove that AB = AC.
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Answer 1

Step-by-step explanation:

quadrilateral ABCD we have

AC = AD

and AB being the bisector of ∠A.

Now, in ΔABC and ΔABD,

AC = AD

[Given]

AB = AB

[Common]

∠CAB = ∠DAB [∴ AB bisects ∠CAD]

∴ Using SAS criteria, we have

ΔABC ≌ ΔABD.

∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.

∴ BC = BD.

Answer 2

Answer:

In △s ABD and ADC,

AD = AD (Common)

BD = CD (Given)

AB = AC (Given)

Hence, △ABD≅△ADC (SSS rule)

Thus, ∠ADB=∠ADC=x (By cpct)

∠ADB+∠ADC=180 (Angles on a straight line)

x+x=180

x=90

∠ADB=∠ADC=90

∘

or, AD is perpendicular to BC