Answer:
Step-by-step explanation:
13. Since, it is given that we have to prove ΔGDH≅ΔFEH by HL rule, thus From ΔGDH, ∠GDH=90° and GH is the hypotenuse and from ΔFEH, ∠HEF=90° and HF is the hypotenuse, thus in ΔGDH and ΔFEH,
∠GDH≅∠HEF=90°
GH≅FH,
Thus by HL the given triangles are congruent.
Option A is correct.
14. From the figure, it can be seen that both triangles have a right angle congruent and hypotenuse congruent, thus by HL rule, triangle scan be prove congruent.
Option A is correct.
15. It is given that BI is parallel to RD, thus using the alternate interior angle property, ∠B≅∠R.
Option C is correct.
16. It is given that BI is parallel to RD, and ∠BSI and ∠RSD form the vertically opposite angles, thus ∠BSI≅∠RSD by vertical angles are congruent.
Option A is correct.
17. Since Δ ABC is an isosceles triangle, thus two sides of the triangle are equal, therefore ∠A=∠B=3x+5, using the angle sum property in ΔABC, we have
3x+5+2x+3x+5=180°
⇒8x+10=180
⇒8x=170
⇒x=21.25
Thus, ∠B=3x+5=3(21.25)+5=63.75+5=68.75°
18. From ΔJKL and ΔDEF, we have
∠K=∠E (given)
∠L=∠F(given)
JL=DF
Thus, By AAS rule ΔJKL ≅ ΔDEF
As corresponding angles and corresponding sides are equal in these two triangles as compared to the other triangles.
, we get 
and 
, so we can parameterize the intersection by
.