Triangles Congruence theorem

**rochosh** · Oct 27th 2012, 05:52 PM

Given two triangles ABC and DEF, if they have one internal angle from each triangle equal, the opposite sides of these(angles)
equal ,and the corresponding height relative to the mentioned side equal. Then they are congruent.

How do we prove it.
I would like to discuss this interesting property and the different methods to tackle this problem.
I've found out two different methods-both of them related to circumferences-that will be posted as soon
as I gather and order the information.

Please let me know if I'm in the wrong topic, it is not my intention to irritate anyone.
Any comment will be taken into account.

**beezmap** · Oct 29th 2012, 04:17 AM

Good question...

**albertjhon** · Nov 4th 2012, 10:02 PM

Good question , I will try this question...

**rochosh** · Feb 28th 2013, 06:47 PM

$Triangles Congruence theorem-tria-1.png$

given:
$\overline{AB}\leq \overline{BC} \and\ \overline{EF}\leq \overline{DE}$

$\overline{AC}=\overline{DF} ; \overline{BH}=\overline{EL} ;\measuredangle ABC = \measuredangle DEF$

The final intention is to prove:
$\triangle ABC\cong \triangle FED$

Theorem:
In any convex quadrilateral ABCD. If any angle( $\theta$ ). made by any of the two diagonal and any side(b), is equal to another angle which is made by the remaining diagonal and the opposite side to b.
then, the it is a tangential quadrilateral.

$Triangles Congruence theorem-qua-1.png$
in order to prove this theorem the triangles
$\triangle ABP \and\ \triangle DCP$ have to be analyzed:
It is true that:
$\measuredangle BPC = \measuredangle APD= \measuredangle CDP + \theta = \measuredangle BAP + \theta \newline \rightarrow\measuredangle CDP = \measuredangle BAP = \beta \newline \rightarrow \measuredangle BPC= \measuredangle APD = \theta + \beta .......(I)\newline$
They have two identical angles, which implies that their angles are all identical
$\rightarrow \triangle BPA \sim \triangle CPD \newline \rightarrow BP/PC=AP/PD......(II)$

If we analyze the triangles:
$\triangle BPC \and\ \triangle APD \newline (I).....\measuredangle BPC= \measuredangle APD \newline (II)..... BP/PC=AP/PD \newline \rightarrow BP/AP=PC/PD$
Two sides have lengths in the same ratio, and the angles included between these sides have the same measure
$\rightarrow \triangle BPC \sim \triangle APD \newline \rightarrow \measuredangle BCP = \measuredangle PDA = \epsilon \newline \rightarrow \measuredangle PBC= \measuredangle PAD =\sigma$

In the quadrilateral, opposite angles add to 180 degrees, thus it is a tangential quadrilateral.

Back to our problem; if the triangles share one more angle measure ,then , they would be similar. But because they have one corresponding side in common, they would be indeed congruent. ...(*)

The next figure shows the triangles with their base sides one over the other, so the points A and D occupy the same position, likewise C and F
$Triangles Congruence theorem-tria-2.png$
If we draw the line BE:
$\overline{BE} \parallel \overline{AC} \newline$
Becouse they are Alternate Interior Angles.
$\rightarrow \measuredangle AEB = \measuredangle CAE \and\ \measuredangle EBC = \measuredangle BCA .......(III)$
$\rightarrow$ Also, The quadrilateral ABEC is tangential......(see theorem above)
$\rightarrow \measuredangle AEB = \measuredangle BCA ....(IV)$
from (III) and (IV), we can conclude that:
$\measuredangle CAE = \measuredangle BCA$ . .....(*)

I will try to post, as soon as possible, a second method(rather intuitive) to prove this problem.

Any comment will be appreciated.

**vicwilson123** · Mar 30th 2013, 11:32 PM

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