Question

ABC is a triangle in which AB=AC and D is a point on AC such that BC^2 = AC ×CD. Prove that BD= BC.

Answer 1

HELLO DEAR,

Given THAT:-

ΔABC, AB = AC

D is a point on AC such that BC² = AC × AD

$\frac{BC}{AD} = \frac{AC}{ BC} .........(1) = > \frac{BC}{CD} = \frac{AC}{ BC}$

In ΔABC and ΔBDC

$\frac{BC}{CD} = \frac{AC}{ BC} \\ = > \: < C = < C \: \: (common)$
ΔABC ~ ΔBDC [By SAS similarity criterion]

$\frac{AB}{BD} = \frac{BC}{CD} = \frac{AC}{ BC}$
$(TRAINGLES ARE SIMILAR CORRESPONDING SIDE ARE PROPORTIONAL)$
$\frac{AC}{BD} = \frac{BC}{CD} \: \: \: \: [AB=AC].....(2)$
From (1) and (2),

we get,

$\frac{AC}{BC} = \frac{AC}{BD}$

BC=CD

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hello everyone...

Hope this helps you....