# In a quadrilateral ABCD ,which is not a trapezium.It is known that <DAB=<ABC=60 DEGREE.moreover , <CAB=<CBD.Then A) AB=BC+CD B) AB=AD+CD C) AB=BC+AD D) AB=AC+AD ? Choose correct option.

Jan 8, 2016

C) $A B = B C + A D$

#### Explanation:

To ease the comprehension of the problem refer to the figure below: If the sides BC and AD are extended the ${\triangle}_{A B F}$, equilateral, is formed.

First, in addition to the angles informed it was possible to determine other interior angles of the triangles ABC and ABD, as shown in the figure.

Then using the Law of Sines on these triangles:

• ${\triangle}_{A B C}$

$\frac{\text{AB"/sin (120^@-alpha)="BC"/sin alpha="AC}}{\sin} {60}^{\circ}$

• ${\triangle}_{A B D}$

$\frac{\text{AB"/sin (60^@+alpha)="AD"/sin (60^@-alpha)="BD}}{\sin} {60}^{\circ}$

To focus on viable hypotheses I suggest to begin trying to prove a specific case.
For instance $A B = 5$ and $\alpha = {15}^{\circ}$:

From ${\triangle}_{A B C} \to \frac{5}{\sin {105}^{\circ}} = \frac{\text{BC}}{\sin} {15}^{\circ}$ => $B C = 5 \cdot \sin {15}^{\circ} / \sin {105}^{\circ}$

From ${\triangle}_{A B D} \to \frac{5}{\sin {75}^{\circ}} = \frac{\text{AD}}{\sin} {45}^{\circ}$ => $A D = 5 \cdot \sin {45}^{\circ} / \sin {75}^{\circ}$

Note that $\sin {75}^{\circ} = \sin 105$ because ${75}^{\circ}$ and ${105}^{\circ}$ are supplementary angles and sines of angles of the first and second quadrants are positive.

Finally,

$A D + B C = 5 \cdot \frac{\sin {15}^{\circ} + \sin {45}^{\circ}}{\sin {75}^{\circ}} = 5 \cdot 1 = A B$

So in this case hypothesis (C) is true (I tried the others and verified that they are false.)

Now for the general proof of the hypothesis (C), following the same steps that worked for the specific proof:

From ${\triangle}_{A B C} \to \frac{\text{AB"/sin (120^@-alpha)="BC}}{\sin} \alpha$

Observation:

$\sin \left({120}^{\circ} - \alpha\right) = \sin \left({180}^{\circ} - \left({120}^{\circ} - \alpha\right)\right) = \sin \left({60}^{\circ} + \alpha\right)$

$\implies \text{AB"/sin (60^@+alpha)="BC"/sin alpha => "BC"="AB} \cdot \sin \frac{\alpha}{\sin} \left({60}^{\circ} + \alpha\right)$

Therefore,

$\text{BC" ="AB} \cdot \frac{\sin \alpha}{\left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha + \left(\frac{1}{2}\right) \cdot \sin \alpha}$

From ${\triangle}_{A B D} \to \frac{\text{AB"/sin (60^@+alpha)="AD}}{\sin} \left({60}^{\circ} - \alpha\right)$

This will get you

$A D = \text{AB} \cdot \sin \frac{{60}^{\circ} - \alpha}{\sin} \left({60}^{\circ} + \alpha\right)$

$= A B \frac{\left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha - \left(\frac{1}{2}\right) \cdot \sin \alpha}{\left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha + \left(\frac{1}{2}\right) \cdot \sin \alpha}$

Finally,

$B C + A D = \text{AB} \frac{\sin \alpha + \left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha - \left(\frac{1}{2}\right) \cdot \sin \alpha}{\left(\frac{\sqrt{3}}{2}\right) \cdot \cos \alpha + \left(\frac{1}{2}\right) \cdot \sin \alpha}$

$B C + A D = \text{AB} \cdot 1$

Therefore,

$A B = B C + A D$ and the hypothesis (C) is true.**

Jun 11, 2016

Option( c)is correct

#### Explanation: Given

• $\text{In quadrilateral } A B C D , \angle D A B = \angle A B C = {60}^{\circ}$

• $\angle C A B = \angle C B D$

Construction

• $B C \mathmr{and} A D \text{ are produced to intersect at E}$

• $I n \Delta A B E , \angle E B A = \angle E A B = {60}^{\circ} \implies \angle B E A = {60}^{\circ}$

• $\therefore \Delta A B E \text{ is equilateral} \to A B = B E = E A$

Now

• $I n \Delta A B C \mathmr{and} \Delta B E D , \angle B E D = \angle C B A = {60}^{\circ}$
$\angle E B D = \angle C A B \left(g i v e n\right) \mathmr{and} B E = A B \left(p r o v e d\right)$
• $\Delta A B C \mathmr{and} \Delta B E D \text{ are congruent by ASA Rule}$
• $\therefore D E = B C \text{ corresponding sides}$
$\text{of congruent triangles}$
• $\text{Finally} , A B = A E = A D + D E = A D + B C$

So option (c) is correct