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

Glencoe/McGraw Hill

1 Answer
Jan 26, 2018

Please see below.

Explanation:

We can use cosine formula for triangles.

We have #RQ=ST#

also #m/_1>m/_2# as #/_1# is exterior angle and is hence sum of #/_2# and #m/_QTS#

and as both angles are less than #180^@#, we have #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#