Question
In Quadrilateral $A B C D$, side $A D \| B C$, diagonal $A C$ and $B D$ intersect in point $P$, then prove that $\frac{ AP }{ PD }=\frac{ PC }{ BP }$
✓
Answer
Proof: seg $A D|| \operatorname{seg} B C$ and $B D$ is their transversal.
...[Given]
$\therefore \angle DBC \cong \angle BDA \quad \ldots$ [Alternate angles]
$\therefore \angle PBC \cong \angle PDA \quad$...(i) $[ D - P - B ]$
In $\triangle P B C$ and $\triangle P D A$,
$\angle P B C \cong \angle P D A \quad \ldots[$ From (i) $]$
$\angle B P C \cong \angle DPA \quad \ldots$ [Vertically opposite angles]
$\therefore \triangle PBC \sim \triangle PDA \quad \ldots[ AA$ test of similarity]
$\therefore \frac{ BP }{ PD }=\frac{ PC }{ AP } \quad \ldots[$ Corresponding sides of similar triangles $]$
$\therefore \frac{ AP }{ PD }=\frac{ PC }{ BP } \quad \ldots[$ By alternendo $]$
...[Given]
$\therefore \angle DBC \cong \angle BDA \quad \ldots$ [Alternate angles]
$\therefore \angle PBC \cong \angle PDA \quad$...(i) $[ D - P - B ]$
In $\triangle P B C$ and $\triangle P D A$,
$\angle P B C \cong \angle P D A \quad \ldots[$ From (i) $]$
$\angle B P C \cong \angle DPA \quad \ldots$ [Vertically opposite angles]
$\therefore \triangle PBC \sim \triangle PDA \quad \ldots[ AA$ test of similarity]
$\therefore \frac{ BP }{ PD }=\frac{ PC }{ AP } \quad \ldots[$ Corresponding sides of similar triangles $]$
$\therefore \frac{ AP }{ PD }=\frac{ PC }{ BP } \quad \ldots[$ By alternendo $]$
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