Calculate the measure of segment AB in triangle rectangle ABC

Question

For reference:

Given the triangle $ABC$, straight at $B$. The perpendicular bisector of $AC$ intersects at $P$ with the angle bisector of the outer angle $B$, then $AF \parallel BP$ ($F\in BC$) is drawn. If $FC$ = $a$, calculate BP(x). (Answer: $\frac{a\sqrt2}{2})$

My progress:

Point P is on the circumcircle of ABC because the angle $\measuredangle ABP = 135^o$ $ Where~ AC = 2R\\ \triangle CBP \rightarrow PC^2 = BP^2 + BC^2 - \sqrt2BCBP\\ but~PC = PA = R\sqrt{2} \text{(since P is in the bisector AC)}\\ 2R^2 = BP^2 +BC^2 - \sqrt{2}BCBP\\ 0= BP^2 - \sqrt{2}BCBP + BC^2 - 2R^2\\ 0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2}2 - 2R^2\\ 0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2-4R^2}2\\ 0= (BP - \frac1{ \sqrt{2}}BC)^2 - \frac{AB^2}2\\ $

I can't find the relationship between BC, AB and a... If anyone finds another way to solve by geometry I would be grateful

Answer 1

$BP \parallel AF \implies ∠PBF= ∠BFA=45° \text{ (alternate interior angles)}$

$ \implies ∠BAF=∠BFA \text{ (sum of angles in triangle)} $

$\implies AB=FB \text{ (sides opposite to equal angles in a triangle)}$

$\therefore BC= CF+FB=a+AB \implies AB=BC-a$

Using this, in the equation you got

$(BP - \frac1{ \sqrt{2}}BC)^2 - \frac{AB^2}2 =0 \implies (BP - \frac1{ \sqrt{2}}BC)^2 - (\frac{BC-a}{\sqrt{2}})^2=0$

$\implies \{(BP - \frac1{ \sqrt{2}}BC)+(\frac{BC-a}{\sqrt{2}})\}\{(BP - \frac1{ \sqrt{2}}BC)-(\frac{BC-a}{\sqrt{2}})\}=0$

$\implies (BP - \frac{a}{ \sqrt{2}})(BP - \frac2{ \sqrt{2}}BC+\frac{a}{ \sqrt{2}})=0$

$\implies BP=\frac{a}{ \sqrt{2}} \text{ or }BP=\sqrt{2} BC-\frac{a}{ \sqrt{2}} \tag{i} \label{i}$

Let us assume $BP=\sqrt{2} BC-\frac{a}{ \sqrt{2}} \tag{ii}$

$$\text{In }\triangle ABF, AB+BF>AF \implies 2(BC-a)>AF (\because AB=BF=BC-a$$

$$\implies 2BC-2a+a>AF+a>AC (\because AF+a>AC \text{ in } \triangle AFC$$

$$\implies \sqrt{2} BC-\frac{a}{ \sqrt{2}}>\frac{AC}{\sqrt{2}} \text{ (divided both sides by }\sqrt{2})$$

$$\implies BP>\frac{AC}{\sqrt{2}}=\frac{2R}{\sqrt{2}}=\sqrt{2}R\text{ (using (ii))}$$

$\implies BP>AP \implies \overset\frown{BP}>\overset\frown{AP}$ which is a contradiction. Therefore our assumption (ii) is wrong.

$\therefore BP\neq \sqrt{2} BC-\frac{a}{ \sqrt{2}}$ and (i) $\implies BP =\frac{a}{ \sqrt{2}} $

$\textbf{Method 2: Using similarity of triangles}$

$\begin{array}{l} \text{In } \triangle ABP \text{ and } \triangle AFC\\ \angle ABP= \angle AFC =135° (\because \angle AFC=180°-\angle BFA)\\ \angle BAP=\angle FAC (\because \angle BAP=\angle BAC - \angle PAC =\angle BAC - 45°=\angle BAC - \angle BAF=\angle FAC) \end{array}$

$\therefore \triangle ABP \sim \triangle AFC$ by AAA similarity criterion.

$\implies \frac{BP}{FC}=\frac{AP}{AC} \implies \frac{BP}{a}=\frac{\sqrt{2} R}{2R}=\frac1{\sqrt{2}}$

$\therefore BP=\frac{a}{\sqrt{2}}$

Answer 2

You have done most of the work by coming up with the nice construction. I will just mark point $Q$, draw chord $CQ$ and use it to find the answer.

As $BP$ and $AQ$ are parallel, $\angle AFB = \angle FBP = 45^0$

So $\angle CFQ = 45^0$ and as $\angle CQF = 90^0$, $CQ = \cfrac{a}{\sqrt2}$

But also note that, $\angle CAQ = \angle BAP = \angle A - 45^0$

So we must have, chord $BP$ = chord $CQ$.

$\therefore x = \cfrac{a}{\sqrt2}$

Answer 3

An uglier brute-force method by some trigonometric identities:

Let the green marked $\angle BCA$ be $\theta$. Then the corresponding angle at centre $\angle BEA = 2\theta$, and $\angle PEB = 90^\circ - 2\theta$.

By the laws of cosine,

$$\begin{align*} BP^2 &= EB^2 + EP^2 - 2 EB\cdot EP \cos \angle PEB\\ &= 2R^2 - 2 R^2 \cos (90^\circ - 2\theta)\\ &= 2R^2 - 2R^2\sin 2\theta\\ &= 2R^2\left(\sin^2\theta -2\sin\theta\cos\theta +\cos^2\theta\right)\\ &= \frac12 (2R)^2\left(\cos\theta-\sin\theta\right)^2\\ &= \frac12 \left[(BF+a)-AB\right]^2\\ &= \frac{a^2}{2} & (AB=BF)\\ x = BP &= \frac{a}{\sqrt2} \end{align*}$$