In the given figure, seg AC and seg BD intersect each other in point P and `"AP"/"CP" = "BP"/"DP"`. Prove that, ∆ABP ~ ∆CDP.
Practice Set 1.3 | Q 8 | Page 22
In the given figure, seg AC and seg BD intersect each other in point P and `”AP”/”CP” = “BP”/”DP”`. Prove that, ∆ABP ~ ∆CDP.
Given: Seg AC and seg BD intersect each other in point P and `"AP"/"CP" = "BP"/"DP"`.
To prove: ∆ABP ~ ∆CDP
Proof: In ∆ABP and ∆DCP,
`"AP"/"CP" = "BP"/"DP"` ...(Given)
∠APB = ∠DPC ...(vertically opposite angles)
By SAS test of similarity,
∆ABP ~ ∆CDP
Hence Proved.
Answer:-
Given: Segments AC and BD intersect at point P, and “AP/CP = BP/DP”.
To prove: ∆ABP ~ ∆CDP. Proof: In ∆ABP and ∆DCP, AP/CP = BP/DP. (Given) ∠APB = ∠DPC (Vertically opposite angles) By SAS test of similarity, ∆ABP ~ ∆CDP. Therefore, the two triangles are similar.
Chapter 1. Similarity- Practice Set 1.3 – Page 22
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