Geometry question for exam:

Given a right isosceles triangle $ABC$ with $A=90^\circ$. Point $D$ is inside triangle $ABC$ and such that $\angle DCA=\angle DBC=30^\circ$. Prove that $DCA$ is an isosceles triangle.

Basicly my idea is to label point $E$ such that $\angle EDC=15^\circ$. And I'm stuck at that point. I have learnt equal triangle but i haven't learnt similar triangle or trigonometry. Thanks for your help!

  • $\begingroup$ The part you have to prove seems to have fallen off your text. What is it? $\endgroup$
    – HSN
    Feb 15, 2017 at 11:51
  • $\begingroup$ It's "Prove that DCA is an isosceles triangle." $\endgroup$ Feb 15, 2017 at 11:53
  • $\begingroup$ Don't delete and re-post you questions. $\endgroup$
    – Asaf Karagila
    Feb 15, 2017 at 13:38

3 Answers 3


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We have that $\angle BDC =180°-45°$ and using sine rule for $\Delta BCD$ we get:

$$\frac{CD}{\sin 30°}=\frac{a\sqrt{2}}{\sin (180°-45°)}=\frac{a\sqrt{2}}{\sin 45°}\to CD=\frac{a\sqrt{2}\cdot \sin 30°}{\sin 45°}=a$$

So $AC=CD=a$ and then $\Delta DCA$ is isosceles.


As Michael pointed out, this property only holds if $D$ lies inside the triangle. But I think that is the case OP is asking for. The statement is incomplete.

  • $\begingroup$ Thanks but I haven't learnt trigonometry yet. $\endgroup$ Feb 15, 2017 at 13:17
  • $\begingroup$ @LêĐứcMinh: what have you learnt so far? $\endgroup$
    – Arnaldo
    Feb 15, 2017 at 13:40
  • $\begingroup$ Just very basic geometry, including equal triangle(I bet we must use it because that is the topic of my test!). I haven't learnt similar triangle or anything about trigonometry or property of quadrilateral of circle.@Arnaldo $\endgroup$ Feb 15, 2017 at 13:44

Problems like this one can usually be solved without trigonometry.

Let $M$ be the midpoint of edge $BC$. Then, since $AB=AC$, line $AM$ is the orthogonal bisector of edge $BC$, i.e. $AM$ is the line orthogonal to $BC$ that passes through $M$. Let $E$ be a point on $AM$ such that $\angle \, EBC = 60^{\circ}$ and both $E$ and $A$ lie on the same side of line $BC$. Since line $EM$ (which is line $AM$) is the orthogonal bisector of $BC$, triangle $BCE$ is isosceles with $BE = CE$.

enter image description here

However, by construction, $\angle \, EBC = 60^{\circ}$ so $BCE$ is in fact an equilateral triangle. Therefore $CE = BC$. Moreover, $EM$ is the angle bisector of angle $\angle \, BEC = 60^{\circ}$ so $\angle \, AEC = 30^{\circ}$. Furthermore, $$\angle \, ACE = \angle \,BCE - \angle \, BCA = 60^{\circ} - 45^{\circ} = 15^{\circ}$$ By the choice of point $D$ $$\angle \, DCB = \angle \, BCA - \angle \, DCA = 45^{\circ} - 30^{\circ} = 15^{\circ}$$ Consequently, triangles $ACE$ and $DCB$ are congruent because $$\angle \, ACE = \angle \, DCB = 15^{\circ}$$ $$\angle \, AEC = \angle \, DBC = 30^{\circ}$$ and $CE = BC$. Therefore $CA = CD$ and thus triangle $DCA$ is isosceles.


It's wrong!

Let $BD\cap AC\equiv\{K\}$, where $BD$ is a segment such that $D$ outside of the triangle.

Thus, $\measuredangle BCD=\measuredangle BDC=75^{\circ}$ and $DC=2a\sqrt2\sin15^{\circ}\neq a$

  • 1
    $\begingroup$ Wait, why is BCD=75? $\endgroup$ Feb 15, 2017 at 13:56
  • $\begingroup$ @Lê Đức Minh $45^{\circ}+30^{\circ}=75^{\circ}$. $\endgroup$ Feb 15, 2017 at 13:59
  • $\begingroup$ Oh, thank you. I missed an information in the problem. $\endgroup$ Feb 15, 2017 at 14:02

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