Question

In the given figure ABC is square and O is a point inside it.such that OD.Prove that AOC is straight line​

Answer 1

Hey let's solve this

Step-by-step explanation:

It is given that ABCD is a square and P is a point inside it such that PB=PD

Considering △APD and △APB

We know that all the sides are equal in a square

So we get DA=AB

AP is common i.e. AP=AP

According to SSS congruence criterion

△APD≅△APB

We get ∠APD=∠APB(c.p.c.t)…(1)

Considering △CPD and △CPB

We know that all the sides are equal in a square

So we get CD=CB

CP is common i.e. CP=CP

According to SSS congruence criterion

△CPD≅△CPB

We get ∠CPD=∠CPB(c.p.c.t)…(2)

By adding both the equation (1) and (2)

∠APD+∠CPD=∠APB+∠CPB…(3)

From the figure we know that the angles surrounding the point P is 360

∘

So we get

∠APD+∠CPD+∠APB+∠CPB=360

∘

By grouping we get

∠APB+∠CPB=360

∘

(∠APD+∠CPD)…(4)

Now by substitution of (4) in (3)

∠APD+∠CPD=360

∘

(∠APD+∠CPD)

On further calculation

2(∠APD+∠CPD)=360

∘

By division we get

∠APD+∠CPD=180

∘

Therefore, it is proved that CPA is a straight line.

Answer 2

Answer:

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