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Numerade Educator



Problem 35 Hard Difficulty

Use these steps to write a paragraph proof of the Hinge Theorem. Given: $\overline{A B} \cong \overline{D E}, \overline{B C} \cong \overline{E F}, \mathrm{m} \angle A B C>\mathrm{m} \angle D E F$ Prove: $A C>D F$
a. Locate $P$ outside $\triangle A B C$ so that $\angle A B P \cong \angle D E F$ and $\overline{B P} \cong \overline{E F} .$ Show that $\triangle A B P \cong \triangle D E F$ and
thus $\overline{A P} \cong \overline{D F}$
b. Locate $Q$ on $\overline{A C}$ so that $\overline{B Q}$ bisects $\angle P B C .$ Draw $\overline{Q P}$. Show that $\triangle B Q P \cong \triangle B Q C$ and thus $\overline{Q P} \cong \overline{Q C}$
c. Justify the statements $A Q+Q P>A P, A Q+Q C=A C, A Q+Q C>A P$ $A C>A P,$ and $A C>D F$


a. $ \overline{A P} \cong \overline{D F}$
b. $\overline{Q P} \cong \overline{Q C}$
c. $A C>D F$


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Video Transcript

were given Triangle ABC in Triangle D E a. Such that segment A B is concluded. The Segment D E and segment BC is congruent to segment eat if we're going to start by contract constructing point B Such that angle. ABP is congruent to angle d E f so we can market this way. And since a B is congrats to D E, that gives us segment, BP is now Congrats to segment E if and, um, since a B is congruent to D E and angle, ABP is congruent to D E f. We now have BP kindred to D E f. So we now have triangle. A BP is congruent to try angle D E f by side angle side congruence the're, um and that allows us to say that Segment A B. I'm sorry. A p is congruent to segment D F by CPI C T. C, which means corresponding parts of congruent triangles are congruent. Next, we wanna find point C on point Q. On segment A. C such that B Q is an angle by sector of segment of angle B p B. C. I'm sorry about that. B que is an angle by sector of angle P B c. So now we have angle P b que is congruent to angle c b que we know that segment. BC is congruent to segment e f and we know that segment. BP has also congruent to segment E. If so, by substitution, we can say segment BC is congratulated to segment BP. We also know that segment B que is Congrats to segment B que by reflexive property of congruence, so we now have triangle B. Q. P is congruent to try angle speak you see by side angle side congruence there, um, and that gives us segment. P. Q. Is congrats to segment CQ by CPC TC Again corresponding parts of congruent triangles. Arkan Groot The last part of our proof comes from using the three sides of a triangle. Our Triangle Inequality statement says that any two sides the sum of any two sides of a triangle must be greater than the third side. So let's go with segment A Q Plus Q P must be greater than segment AP That's in our new triangle that we created, but given Point Q. His on segment A C. Our segment addition postulate states that. Oh, excuse me. A que plus Q C must equal a c. We have proven that segment P Q. Is Congrats it to segment que si so by substitution, we now have a que plus. Q C is greater than AP and by substitution. If a Q plus Q C is equal to a C, we could say A C is greater than a P. And before we proved that segment, AP was congruent to segment D f. We did that previously, so we now can substitute that and say that a C must be greater than D. F.

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