This page shows how to construct a
right triangle that has both the given leg lengths.
It is possible to draw more than one triangle has the side lengths as given.
You can use the triangle to the left or right of the initial perpendicular, and also draw them below the initial line.
All four are correct in that they satisfy the requirements, and are
congruent to each other.
This construction works by effectively building two congruent triangles.
The image below is the final drawing above with the blue lines PQ and QA added
Argument
Reason
We first prove that ∆BCA is a right triangle
1
CP is
congruent to CA
They were both drawn with the same compass width
2
PQ is
congruent to AQ
They were both drawn with the same compass width
3
CQ is common to both triangles ∆PQC and ∆AQC
Common side
5
∠QCP, ∠QCA are
congruent
CPCTC. Corresponding parts of congruent triangles are congruent
6
m∠QCA = 90°
∠QCA and ∠QCP are a
linear pair and (so add to 180°)
and
congruent so each must be 90°
7
∆BCA is a right triangle
∠BCA = 90°.
We now prove the triangle is the right size
9
BC is
congruent to the given leg L2
Drawn with same compass width
10
∆BCA is a right triangle with the desired side lengths
(7), (8), (9)
-
Q.E.D
containing two LL triangle construction problems.
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