Find $\angle ABC$ in the triangle below,

Question

In a $\triangle ABC$, draw the heights $AP$ and $CQ$ and the interior angle bisector $BD$. Calculate $ m \angle ABC$, knowing that:$\dfrac{1}{AP}+\dfrac{1}{CQ} = \dfrac{2}{BD}$ (Answer provided: $30^o$)

I try:

$\dfrac{AD}{AB} = \dfrac{CD}{BC}$

$AC^2=AQ^2+CQ^2 \implies CQ^2 = AC^2-AQ^2$

$BC^2 = CQ^2+BQ^2 \implies BC^2 = AC^2-AQ^2+BQ^2$

$AB^2 = AD^2+BD^2 \implies BD^2=AB^2-AD^2$

$BC^2=BD^2+CD^2 \implies BC^2 = AB^2-AD^2+CD^2$

Answer 1

Set $a=BC, b=AC, c=AB, 2x =\angle ABC$. Use the law of sines in triangle $BAD$

$$\frac{A D}{\sin x}=\frac{B D}{\sin A}=\frac{B D}{C Q/b}$$

Similarly, in triangle $BCD$

$$\frac{C D}{\sin x}=\frac{B D}{\sin C}=\frac{B D}{A P/b}$$

Add

$$\frac{b}{\sin x}=\frac{\text{AD} + \text{CD}}{\sin x}=\frac{A D}{\sin x}+\frac{C D}{\sin x}=\frac{B D}{C Q/b}+\frac{B D}{A P/b}= BD \left( \frac{b}{\text{CQ}}+\frac{b}{\text{AP}}\right)=2 b $$

$$\sin x=\frac{1}{2}$$

$$x=30^{\circ} $$

$$\angle ABC = 60^{\circ}$$

Answer 2

Comment: In equilateral triangle we have:

$\frac1 s+\frac 1 s=\frac 2s$

where s is the area of triangle. We write this relation in terms of altitudes and sides:

$\frac 1 {ah_a}+\frac 1 {ch_c}=\frac 2 {bh_b}$

$a=b=c$

therefore:

$\frac 1{h_a}+\frac 1{h_c}=\frac 2{h_b}$

which is given relation in statement if BD is also the altitude of the triangle.. Hence triangle must be equilateral and we must have $\angle ABC=60^o$

Answer 3

Continuing on my comment on sirious erroneous answer, the triangle in OP must not be equilateral.

In fact the correct answer above by Lozenges shows that a necessary and sufficient condition is that the bisected angle measure is $60^\circ$.

To further emphasize this point, we can easily prove that, provided that $\measuredangle ABC = 60^\circ$, the internal bisector of this angle is the harmonic mean of the altitudes w.r.t. sides $AB$ and $BC$ (the converse of what stated by OP). Consider the figure below, where $CS\parallel BN$ and $BT\perp CS$.

Let now $\measuredangle ABC = 60^\circ$, $\overline{AH} = x$, $\overline{CK} = y$ and $\overline{BN} = z$.

We then have $\overline{AB} = \frac{2}{\sqrt 3}x$, and $\overline{BC} = \frac{2}{\sqrt 3} y$.

Since $\triangle BCS$ is isosceles with $\measuredangle CBS = 120^\circ$, we also have $\overline{BS} = \frac{2}{\sqrt 3} y$, and $\overline{CS} = 2y$.

Similarity $\triangle ABN \sim \triangle ASC$ yields $$\frac{z}{2y}=\frac{\frac{2}{\sqrt 3}x}{\frac{2}{\sqrt 3}x+\frac{2}{\sqrt 3}y}$$ which is equivalent to $$\frac{2}{z} = \frac1x + \frac1y.$$

(In the figure above I took $\overline{AH} = 2$ and $\overline{BK}= 3$, so that $\overline{BN} = \frac{12}5$.)

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As a final note, the above construction can be used, of course, to give an alternative proof of OP, without trigonometric expressions, as follows.

Suppose we have $$\frac{2}{z} = \frac1x+\frac1y.\tag{1}\label{1}$$

1. Similarity $\triangle ABH \sim \triangle BCK$ gives $\overline{AB} = kx$ and $\overline{BC} = ky$, for some positive $k$.
2. The fact that $\triangle CBS$ is isosceles and similarity $\triangle ABN \sim \triangle ACS$ yields $$\frac{z}{kx} = \frac{\overline{CS}}{kx+ky},$$ that is $$z = \frac{x\cdot \overline{CS}}{x+y}.\tag{2}\label{2}$$
3. From \eqref{1} we have $$z = \frac{2xy}{x+y}\tag{3}\label{3}.$$
4. Comparing \eqref{2} and \eqref{3} leads to $$\overline{CS} = 2y,$$ that is $$\overline{CT} = y.$$
5. By HL congruence $\triangle BCK \cong \triangle BCT$. Hence $$\angle ABC \cong \angle CBT \cong \angle TBS.$$ Therefore $$\boxed{\measuredangle ABC = 60^\circ}.$$

Answer 4

In the figure, draw $DX$, $DY$ perpendicular to $BC$ and $BA$ as shown.

$(1) \Delta BDX \cong \Delta BDY \implies DX=DY.$

$(2) \Delta ADY \sim \Delta ACQ \implies \frac{DY}{CQ}=\frac{AD}{AC}.$

$(3) \Delta CDX \sim \Delta CAP \implies \frac{DX}{AP}= \frac{CD}{AC}.$

$(4)$ From $(1), (2), (3)$, we have $$\frac{DY}{CQ}+\frac{DX}{AP}=\frac{AD}{AC}+\frac{CD}{AC}$$ $$\frac{DX}{CQ}+\frac{DX}{AP}=\frac{AD+CD}{AC}$$ $$\frac{DX}{CQ}+\frac{DX}{AP}=1$$ $$\frac{1}{CQ}+\frac{1}{AP}=\frac{1}{DX}$$

$(5)$ Since it is given that $$\frac{1}{CQ}+\frac{1}{AP}=\frac{2}{BD}$$

$(6)$ $$\therefore \frac{1}{DX}=\frac{2}{BD}$$ $$\frac{DX}{BD}=\frac{1}{2}$$ $$\sin \left( \frac{B}{2} \right)=\frac{1}{2}$$ $$\frac{B}{2}=30^{\text o}$$ $$B=60^{\text o}$$