How to prove this property of parabola using geometry?

Question

Tangents are drawn from a point A to a parabola. These tangents touch the parabola at points B and C. Now, if we draw a tangent to the parabola on a point D (on the parabola) such that it intersects AB at P and AC at Q. Then

$$\frac{\mathrm{AP}}{\mathrm{AB}}+\frac{\mathrm{AQ}}{\mathrm{AC}}=1$$

I proved this using analytical approach (Taking parabola $y^2=4ax$, and other points accordingly) and succeeded. But I tried lot but couldn't prove this geometrically. Can you please give me hint on how to approach this proof geometrically.

I am currently working and have proceeded this much.

$$\frac{\mathrm{AP}}{\mathrm{AB}}+\frac{\mathrm{AQ}}{\mathrm{AC}}= \frac{\mathrm{AP \cdot AC}+\mathrm{AB \cdot AQ}}{\mathrm{AB \cdot AC}}=\frac{\mathrm{\dfrac{1}{2}\sin ({A})\cdot AP \cdot AC}+\mathrm{\dfrac{1}{2}\sin ({A}) \cdot AB \cdot AQ}}{\mathrm{\dfrac{1}{2}\sin ({A}) \cdot AB \cdot AC}} =\frac{\operatorname {Ar.}(\Delta APC)+\operatorname {Ar.}(\Delta AQB)}{\operatorname {Ar.}(\Delta ABC)}$$

Therefore, what remains to prove is

$$\operatorname {Ar.}(\Delta APC)+\operatorname {Ar.}(\Delta AQB)=\operatorname {Ar.}(\Delta ABC)$$ Which is same as proving $$\operatorname {Ar.}(\Delta AQB)=\operatorname {Ar.}(\Delta BPC)$$

Thanks.

Answer 1

This question was answered in the comments. The comments can be paraphrased as:

Apollonius' Theorem at cut-the-knot.org establishes $\dfrac{AP}{PB} = \dfrac{QC}{AQ}$.

From this we can show that $$\frac{AP}{AB} = \frac{QC}{AC} =\frac{AC-AQ}{AC} =1- \frac{AQ}{AC}.$$ – Hari Shankar Sep 1 '17 at 10:35

An alternate link for this theorem by Apollonius is Dorrie's 100 Great Problems Of Elementary Mathematics, pg 220.