Two similar isosceles triangles are constructed outside of an parallelogram ABCD, the first being $ABB_1$ and second $CBC_1$ i.e. $|AB| = |AB_1|$ and $|CB| = |CC_1|$. Since $ABB_1$ and $CBC_1$ are similar, prove that the triangle $DB_1C_1$ is similar to $ABB_1$ and $CBC_1$.

I don't know where to start, I don't see any relations so it would be great if someone could give me a starting point/hint what to look for so that I can try to solve the rest by myself.

As for now, I am only allowed to use elementary geometry i.e. Thales, Pythagoras, theorems of the similarity/coherence of 2 triangles, the surface of triangles and quadrilaterals, theorems for circles (the angle over an arc) and the properties of tangential quadrilaterals and cyclic quadrilaterals but no trigonometry properties (we don't use any trigonometry as for now).

I tried to prove the similarity but I am missing an angle or a side to use a similarity theorem.

If I understood the task correctly, the picture should look something like this


and by looking at the triangles $DAB_1$ and $DCC_1$ I notice: ABCD is a parallelogram and $BCC_1$ is an isosceles triangle so $|AD| = |BC| = |CC_1|$.

The angles $DAB_1$ and $DCC_1$ are equal because $$ \angle DAB_1 = \angle DAB + \angle B_1AB $$ $$ \angle DCC_1 = \angle DCB + \angle C_1CB $$ $$ \angle DAB = \angle DCB$$ because ABCD is a parallelogram $$ \angle B_1AB = \angle C_1CB $$ because they are both angles within similar triangles

but I am missing an information (angle or side) which I don't know how to get/conclude because the task says that I have to prove that the red triangle ( $DB_1C_1$ ) is similar i.e. isosceles. So I can't conclude that both triangles ($DAB_1$ and $DCC_1$) have another side of the same size.

ABCD is a parallelogram so $|DC| = |AB| $ and $|AB| = |AB_1| $ because the triangle $ABB_1$ is isosceles hence $|DC| = |AB_1|$

Now I got 2 sides and an angle between them.

I will make another update later and let you know if I need more help (or hopefully have figured it out).

In update 1 and 2 I concluded the similarity and (since the sides are of the same size) coherency of the triangles $DAB_1$ and $DCC_1$ and since they are equal, the sides $DB_1$ and $DC_1$ of the triangle $DB_1C_1$ are equal which proves that the triangle $DB_1C_1$ is an isosceles triangle.

Last step, the angles:

As for now, we have:
$\alpha = \angle DAB = \angle DCB$
$\beta = \angle BAB_1 = \angle BCC_1$

Now using the similarity of triangles $DAB_1$ and $DCC_1$:
$\gamma = \angle DB_1A \implies \gamma = \angle CDC_1$
$\delta = \angle B_1DA \implies \delta = \angle DC_1C$

Now, our notation on the triangle $DAB_1$ implies: $\alpha + \beta + \gamma + \ \delta = 180°$ $\implies \beta = 180° - \alpha - \gamma - \delta \implies \beta = \angle ADC - \gamma - \delta \implies \beta = \angle B_1DC_1$

Note: $\alpha = \angle DAB \implies 180° - \alpha = \angle ADC$ (ABCD is a parallelogram)

And that's our needed angle (because we already proved that the triangle $B_1DC_1$ is an isosceles triangle $\implies \angle DB_1C_1 = \angle DC_1B_1 = \frac{1}{2} (180° - \beta)$ )

$\implies ABB_1 \sim CBC_1 \sim DB_1C_1$


Hint: Prove that triangles $DAB_1, C_1CD$ and $C_1BB_1$ are similar.

  • $\begingroup$ I updated my post with the final steps. Thank you very much!!! $\endgroup$ – AltairAC Mar 26 '13 at 21:38
  • $\begingroup$ @Shirohige Well done! $\endgroup$ – Ivan Loh Mar 27 '13 at 0:14

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