I am having trouble solving this Hinge Theorem Geometry proof. I understand the Hinge Theorem, but how is m<1 congruent to m<2?

Question

2857 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-26T17:04:36

Please see below.

We can use cosine formula for triangles.

We have $R Q = S T$ $RQ=ST$

also $m ∠ 1 > m ∠ 2$ $m/_1>m/_2$ as $∠ 1$ $/_1$ is exterior angle and is hence sum of $∠ 2$ $/_2$ and $m ∠ Q T S$ $m/_QTS$

and as both angles are less than $180^{\circ}$ $180^@$ , we have $cos (∠ 1) < cos (∠ 2)$ $cos(/_1)< cos(/_2)$

Now using cosine formula in $DeltaQRS$ , we have

$RS^2=RQ^2+QS^2-2RQxxQSxxcos(/_1)$ ..................(A)

and in $DeltaQST$ , we have

$QT^2=ST^2+QS^2-2STxxQSxxcos(/_2)$ ..................(B)

Compare RHS of (A) and (B),

we have $RQ=ST$ , but as $cos(/_1)< cos(/_2)$

$RS^2 > QT^2$

i.e. $RS>QT$