In △ ABC, ab = AC, ad is the middle line, P is the upper point of AD, CF ∥ AB is made through point C, and the extension BP intersects AC at point E.1, which proves that BP = PC 2, and the square of BP = PE × pf?

In △ ABC, ab = AC, ad is the middle line, P is the upper point of AD, CF ∥ AB is made through point C, and the extension BP intersects AC at point E.1, which proves that BP = PC 2, and the square of BP = PE × pf?

∵ △ ABC is isosceles triangle, ad is midline,
∴BP=CP,∠ABP=∠ACP
∵AB‖CF ∴∠ABP=∠F
The common angle is ∫ f = ∠ ACP ∫ EPC
∴△PCE∽△PCF
∴PC/PF=PE/PC∴PC²=PF×PE
∵BP=CP ∴BD²=PF×PE

In △ ABC, ab = AC, ad ⊥ BC, CF ∥ AB, BF intersect ad with P, AC with E

Connect PC, ∵ AB = AC, ad ⊥ BC, ∵ ad is the vertical bisector of BC, Pb = PC; there are ∵ PBC = PCB; ∵ AB = AC, ∵ ABC = ACB; there are ∵ ABP = ACP; ∵ CF ∥ AB, ∵ ABP = PFC = ACP

In the known triangle ABC, AB equals AC, ad is the middle line on the edge of BC, CF / / BA, BF intersect ad at point P, AC at point E, and prove that the square of B times P equals PE times PF

The conclusion of connecting CP △ CPE and △ FCP is true ∠ ECP = ∠ ABP = ∠ CFP (AD is the middle line on the edge of BC, ab = AC AP common ∠ bad = ∠ CAD ≌ APB ≌ △ APC ab ‖ CF 〉) ∠ PEC = ∠ a + ABF = ∠ ACP + ECF = ∠ PCF ℅ △ PEC ≌ PCF ‖ BP & # 178; = PE × PF

As shown in the figure, in the triangle ABC, AB equals AC and ad is the middle line. The square of Pb equals PE times PF
Verification of CF parallel ab
I can't send pictures on my tablet,

From ab = AC, ad is the middle line
PB=PC
PC²=PE*PF
PC/PE=PF/PC
△PCE∽△FPC
∠PCE=∠F
You can prove ∠ PCE = ∠ PBA by yourself
So parallel