Mathematics for the first year of junior high school, the properties of isosceles triangles, how to

Answer to question 9.

Take a good look at you, I hope it helps you, ah,

Evidence triangle AHE and triangle BCE congruence (corner corners).

Because CBA=90°, AB=BC, A= C=45°, because EF=FC, angle C=45°, so CFE=90°, so AF=90°, because A=45°, AF=DF

ef=fc

Then EFC is an isosceles triangle.

If abc is an isosceles right triangle, then fec= c=45° then bed=45°

then d=45°, and a=45°, so afd=90°, so afd is an isosceles right triangle.

then af=df

The title is incorrect. It should be abc=90

Your topic is wrong: correct the circumference of an isosceles orange triangle that is known to be 20cm. If one of the edges is 6cm.

Find the length of the other two sides? Suppose this side is waist length, then, the other waist is 6cm, and the third side = 20-6-6=8cm Answer: 6 and 8 If this side is the bottom, then the waist length is (20-6) 2=7cm Answer: 7 and 7

Let the waist length be 2x, and the bottom edge should be yif 2x

x=9xy=8.

x=3,y=5

The waist length is 6, and the bottom edge is 5

Match both sides and greater than the third side if 2x

x=8xy=9.

x=8/3,y=19/3

The waist length is 16 3, and the bottom edge is 19 3

Conform to both sides and be greater than the third side.

It is known that ap=cp, 1= c

AEP CFP (Corner Corner).

PE=PF triangle PEF is always an isosceles right triangle.

Proof: Extend FP and make BK AC to cross FP at point K, connect EECF BK Pbk = PCF

pfc=∠pkb →△bpk≌△cfp→kp=fp→pb=pc

kp=fp →

ek=fe，∠ekp=∠efp=θ，∠kep=∠fep=γep⊥kf →

In the quadrilateral BKFA, it is known that A= ABK=90° has PFA+ PEA=180°= PEA+ PEB= PEB+ PKB

i.e. peb= pfa

pea=∠pkb

and epk= epf=90°, then the quadrilateral pkbe has an external circle a, and the quadrilateral pfae also has an external circle b

and a diameter ek = b diameter ef

i.e. there is a quadrilateral pfae quadrilateral pfae

According to the corner correspondence, there is PE=PF

Give some face, it's good to be a friend.

Proof: Extend FP and make BK AC to cross FP at point K, connect EECF BK Pbk = PCF

pfc=∠pkb →△bpk≌△cfp→kp=fp→pb=pc

kp=fp →

ek=fe，∠ekp=∠efp=θ，∠kep=∠fep=γep⊥kf →

In the quadrilateral BKFA, it is known that A= ABK=90° has PFA+ PEA=180°= PEA+ PEB= PEB+ PKB

i.e. peb= pfa

pea=∠pkb

and epk= epf=90°, then the quadrilateral pkbe has an external circle a, and the quadrilateral pfae also has an external circle b

and a diameter ek = b diameter ef

i.e. there is a quadrilateral pfae quadrilateral pfae

According to the corner correspondence, there is PE=PF

Connect the AP. It is easy to know ap bc, and it is known that epf=90°, it is known, ape= cpf, it is easy to know pae= c=45°, pa=cp, aep cfp, and thus pe=pf