Question

21
In a AABC, AB = AC and D is a point on AC such that
BG? = AC X DC. Prove that BD = BC.​

Answer 1

Answer:

can check below↓↓

Step-by-step explanation:

Given: A △ABC in which AB = AC. D is a point on AC such that BC2 = AC × CD.

To prove : BD = BC

Proof : Since BC2 = AC × CD

Therefore BC × BC = AC × CD

AC/BC = BC/CD .......(i)

Also ∠ACB = ∠BCD

Since △ABC ~ △BDC [By SAS Axiom of similar triangles]

AB/AC = BD/BC ........(ii)

But AB = AC (Given) .........(iii)

From (i),(ii) and (iii) we get

BD = BC.

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Answer 2

Answer:

Answer:

Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the ground

Height and Distance mcq solution image

∴ In ∆ABC, ∠B = 90°

Let height of wall AB = h

Then

sinθ=ABAC⇒sin60∘=h15

⇒3–√2=h15⇒h=153–√2m

∴ Height of the wall =153–√m