Question

ABC and DBC are two isosceles triangle on the same base BC and vertices A and D are on the same side of BC. If ad is extended to introduce AC BC at P show that
1. ∆ABD~= ∆ACD
2. ∆ABD~=∆ACP
3. AP bisect Angle A as well as Angle D
4. AP is the perpendicular bisector of BC​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

1. ABD ~= ACD

AB = AC ( ABC is an isoceles triangle )

DB = DC ( DBC is an isoceles triangle)

AD = AD (common side )

Thus, by SSS congruence rule, ABD ~= ACD.

2. Is the question right?

Is it ABP ~= ACP

3. Since ABP ~= ACP, their corresponding sides are equal. Thus, <BAD= <CAD

4. Since ABP ~= ACP, their corresponding sides are equal. Thus,

<BPA = <CPA. ---eq. 1

It is clear from the figure that,

<BPC = 180°-----eq.2

Combining eq. 1 and eq. 2, you get

<BPA = <CPA = 90°.

Thus, AP is the perpendicular bisector.

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Hope it helps...!

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