Question
Two right triangles $\text{ABC}$ and $\text{DBC}$ are drawn on the same hypotenuse $B C$ and on the same side of $B C$. If $A C$ and $B D$ intersect at $P$, prove that : $A P \times P C=B P \times D P .$
✓
Answer
$\triangle A B C$ and $\triangle D B C$ are right angled triangles at $A$ and $D$ respectively.
$\triangle D P C$ is a right angled triangle at $D$.
Represent the given information as a diagram,
Using Pythagoras Theorem,
$C P^2=C D^2+D P^2$
$\Rightarrow C D^2=C P^2-D P^2$
Now, $C P=C A+A P$ and $D P=D B+B P$
$\Rightarrow C D^2=(C A+A P)^2-(D B+B P)^2$
$\text { Use }(a+b)^2=a^2+2 a b+b^2$
$\Rightarrow C D^2=C A^2+2 C A \cdot A P+A P^2$
$\left(D B^2+2 D B \cdot B P+B P^2\right)$
$\Rightarrow C D^2=C A^2+A P^2+2$
$C A \cdot A P-D B^2-B P^2-2 D B \cdot B P$
$\Rightarrow C D^2+D B^2=C A^2-\left(B P^2-A P^2\right) +2 C A A P-2 D B \cdot B P$
$\Rightarrow C B^2=C A^2-A B^2+2 C A \cdot A P-2 D B \cdot B P$
$\Rightarrow 2 A B^2=2 C A \cdot A P-2 D B \cdot B P$
$\left(C B^2=A B^2+A C^2 \Rightarrow C B^2-A C^2=A B^2\right)$
$\Rightarrow A B^2=(P C-A P) A P-(D P-B P) B P$
$\Rightarrow A B^2=A P \cdot P C-A P^2-D P \cdot B P+B P^2$
$\Rightarrow A B^2=A P \cdot P C-D P \cdot B P+A B^2$
$\Rightarrow A P \times P C=D P \times B P$
Hence proved.
$\triangle D P C$ is a right angled triangle at $D$.
Represent the given information as a diagram,
Using Pythagoras Theorem,
$C P^2=C D^2+D P^2$
$\Rightarrow C D^2=C P^2-D P^2$
Now, $C P=C A+A P$ and $D P=D B+B P$
$\Rightarrow C D^2=(C A+A P)^2-(D B+B P)^2$
$\text { Use }(a+b)^2=a^2+2 a b+b^2$
$\Rightarrow C D^2=C A^2+2 C A \cdot A P+A P^2$
$\left(D B^2+2 D B \cdot B P+B P^2\right)$
$\Rightarrow C D^2=C A^2+A P^2+2$
$C A \cdot A P-D B^2-B P^2-2 D B \cdot B P$
$\Rightarrow C D^2+D B^2=C A^2-\left(B P^2-A P^2\right) +2 C A A P-2 D B \cdot B P$
$\Rightarrow C B^2=C A^2-A B^2+2 C A \cdot A P-2 D B \cdot B P$
$\Rightarrow 2 A B^2=2 C A \cdot A P-2 D B \cdot B P$
$\left(C B^2=A B^2+A C^2 \Rightarrow C B^2-A C^2=A B^2\right)$
$\Rightarrow A B^2=(P C-A P) A P-(D P-B P) B P$
$\Rightarrow A B^2=A P \cdot P C-A P^2-D P \cdot B P+B P^2$
$\Rightarrow A B^2=A P \cdot P C-D P \cdot B P+A B^2$
$\Rightarrow A P \times P C=D P \times B P$
Hence proved.
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