In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that [frac{AP}{PD} = frac{PC}{BP}]
Problem Set 1 | Q 11 | Page 29
In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that [frac{AP}{PD} = frac{PC}{BP}]
Solution
Given: ▢ABCD is a parallelogram
To prove: [frac{AP}{PD} = frac{PC}{BP}]
Proof: In △APD and △CPB
∠APD = ∠CPB (Vertically opposite angles)
∠PAD = ∠PCB (Alternate angles, AD || BC and BD is a transversal line)
By AA test of similarity
△APD ∼ △CPB
[therefore frac{AP}{PC} = frac{PD}{PB} left( text{ Corresponding sides are proportional } right)]
[ Rightarrow frac{AP}{PD} = frac{PC}{PB}]
Hence proved.
Answer:_
Given: ▢ABCD is a parallelogram
To prove: [frac{AP}{PD} = frac{PC}{BP}]
Proof:
Consider △APD and △CPB.
We have,
- ∠APD = ∠CPB (Vertically opposite angles)
- ∠PAD = ∠PCB (Alternate angles, AD || BC and BD is a transversal line)
By AA test of similarity, we can say that △APD ∼ △CPB.
Therefore, we can conclude that the corresponding sides of these two triangles are proportional, which gives us:
[frac{AP}{PC} = frac{PD}{PB}]
Multiplying both sides by PD/PB, we get:
[frac{AP}{PD} = frac{PC}{PB}]
Hence, the proof is complete.
Problem Set 1 | Q 11 | Page 29
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