Circles are fun

Relevant wiki: Cyclic Quadrilaterals

Let $PD=x$ . By the properties of cyclic quadrilaterals we get $\angle BPD=\angle CPD=60^\circ$ and $\angle BPC=120^\circ$ . We have $[BPD]+[CPD]=[BPC]$ , which is:

$\dfrac{21x\sin 60^\circ}{2}+\dfrac{28x\sin 60^\circ}{2}=\dfrac{21(28)\sin 120^\circ}{2}$

$x=\dfrac{21\times 28}{21+28}=\boxed{12}$ .

Draw the diagram and use the fact that angles standing on a chord are equal to identify two pairs of similar triangles.

Since $\triangle DPC$ is similar to $\triangle DBA$ we can say

$\frac{DB}{DP}=\frac{BA}{PC}$

and since $\triangle DPB$ is similar to $\triangle DCA$ we can say

$\frac{DC}{DP}=\frac{CA}{PB}$

Adding these two equations gives

$\frac{DB+DC}{DP}=\frac{BA}{PC}+\frac{CA}{PB}$

Now notice that each numerator equals the side length of the equilateral triangle! Dividing by this gives

$\frac{1}{DP}=\frac{1}{PC}+\frac{1}{PB}=\frac{PC+PB}{PC\times PB}$

and so

$DP=\frac{PC\times PB}{PC+PB}=\frac{21\times 28}{21+28}=\frac{21\times 28}{49}=3\times 4=\boxed{12}$

We shall first prove that $AP=BP+CP=49$ . It is clear that $AP>CP$ . Construct a point $E$ on $AP$ such that $AE=CP=28$ . Then, it suffices to show that $BP=EP$ . Since $AB=BC,AE=CP$ and $\angle BAP=\angle BCP$ , therefore $\triangle ABE\equiv \triangle CBP$ . Thus, $BP=BE$ and $\angle ABE = \angle CBP$ . For $\triangle BPE$ , $\angle EBP=\angle ABC=60^\circ$ . Thus, $\triangle EBM$ is equilateral, so $BP=EP$ and the conclusion is proven.

Now, we find that $\angle BAP=\angle BCP\\\angle ADB=\angle CDP\\\angle ABD=\angle CPD\\\Rightarrow\triangle ABD\sim\triangle CPD\\\Rightarrow \frac{AD}{DC}=\frac{DB}{DP}=\frac{AB}{PC}\\\Rightarrow\frac{AD}{DC}=\frac{BC-DC}{49-AD}=\frac{BC}{28}$

Similarly, $\frac{BC}{21}=\frac{DC}{49-AD}=\frac{AD}{BC-DC}$

Solving these two equation, we get $AD=37$ . Therefore, $PD=49-37=12$ . The answer is thus 12.

Let $PD=x, AB=AC=BC=a, CD=y, BD=a-y$ arc PC subtends 2 equal angles so $\angle PBC=\angle PAC$ and similarly $\angle PCB=\angle PAB$ . So $\triangle PBD\sim\triangle CAD$ (opp angles & angles above are equal) $\implies \frac{21}{a}=\frac{x}{y}$ and similarly $\triangle PCD\sim\triangle BAD \implies \frac{28}{a}=\frac{x}{a-y}$ Cross-multiplying each eqn. And eliminating $ax$ gives $21y=28(a-y) \implies y=\frac{4}{7}a$ . From eqn 1, $x=\frac{21y}{a}=\dfrac{21(\frac{4}{7}a)}{a}=\boxed{12}$

Angles by chords AB and AC are 60, so BPD=CPD=60 and so BPC=120.
Triangle BPC:-
$Using\ Cos\ Rule\ BC=\sqrt{21^2+28^2- 2*21*28*CosBPC}=42.579.$
So s=1/2 21 28*42.579=45.579.
Since PD is the angle bisector of BPC,
$PD=\frac{21*28*45.578967*(45.579-42.579)}{21+28}= \color{#D61F06}{12}$

Imgur Angle bisector PD divides the base BC in 21:28 = 3:4. Let these segments be 3x and 4x

Let, d = PD the length to be found

From Stewart's Theorem, $21^2 \times 4x + 28^2 \times 3x = 7x \times d^2 + 84 x^2$ giving $d^2 + 12 x^2 = 588$ . . . .(A)

Using cosine rule for 120° angle BPC gives: $-\frac{1}{2} = \frac{21^2 + 28^2 - 49 x^2}{2 \times 21 \times 28}$ giving $x^2 = \frac{1813}{49}$ . . . (B)

From (A) and (B) $\textbf{ d = 12}$